Question

Ka for HCN is 4.9×10−10 and Kb for NH3 is 1.8×10−5, calculate Kb for CN− and Ka for NH4+. Enter the Kb value for CN− followed by the Ka value for NH4+, separated by a comma, using two significant figures.

Answer 1

We know that K_a x k_b = k_w

(a) Given K_a for HCN is K_a = 4.9×10⁻¹⁰

                HCN + H₂O CN^- + H₃O⁺

   k_w = hydrolysis constant of water = 1.0x10 ^-14

So K_b for CN^- is k_b = k_w / k_a

                                = ( 1.0x10 ^-14 ) / (4.9×10⁻¹⁰ )

                                = 2.0 x10 ^-5

Therefore the K_b for CN^- is 2.0 x10 ^-5

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Given K_b for NH₃ is 1.8×10⁻⁵

   NH₃ + H₂O NH₄⁺ + OH^-

So K_a for NH₄⁺ is k_a = k_w / k_b

                                = ( 1.0x10 ^-14 ) / (1.8×10⁻⁵)

                                = 2.0 x10 ^-5

Therefore the K_a for NH₄⁺ is 5.6 x10 ^-10

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Therefore the required format is : 2.0 x10 ^-5 , 5.6 x10 ^-10